Aakash Repeater Courses
ApplyTake Aakash iACST and get instant scholarship on coaching programs.
NCERT Exemplar Class 11 Maths Solutions chapter 5 covers Complex numbers and their applications. In our daily lives, we come across many concepts involving real and imaginary components, such as electrical circuits and signal processing. So, what is a complex number? A complex number is a number that has two parts: a real part and an imaginary part. It is written in the form a + bi, where a is the real part and b is the imaginary part with i² = -1. Complex numbers are useful in solving equations that have no real solutions. We have been studying complex numbers over the years, with any ‘real number’ being said to be a complex number, which can be denoted as a variable or alphabet in simpler terms. The other number, along with the real number, is a part of the complex number, and is called the ‘imaginary number.’ Both of them appear together in a problem.
Terms such as integer, conjugate, square root, polar, etc. will be studied in NCERT Exemplar Class 11 Maths Solutions Chapter 5, which will be utilised to find results for these complex number problems. Unchanging practice NCERT Solutions for Class 11 via a worksheet and exercise is highly recommended for students preparing for a tough examination, as it contributes to a deep understanding of the subject and enables them to perform analogous tests.
Class 11 Maths chapter 5 solutions Exercise: 5.3 Page number: 91-97 Total questions: 50 |
Question:1 For a positive integer n, find the value of $(1 - i)^n$ $(1-1/i)^n$
Answer:
$(1-i)^n (1-1/i)^n$
$=(1-i)^n (1+i )^n$
$=(1-i^2 )^n$
$=2^n$
Question:2 Evaluate $\sum_{i=1}^{13}$$(i^n+i^{n+1} )$ where n $\epsilon$N.
Answer:
$\sum_{i=1}^{13}$$(i^n+i^{n+1} )$
$=$ $\sum_{i=1}^{13}$$(1+i) i^n$
$=(1+i)(1+i^2+i^3+i^4....+i^{12}+i^{13} )$
$=(1+i) \frac{i(i^{13}-1)}{(i-1)}$
$=(1+i) i\frac{(i-1)}{(i-1)}$
$=(1+i)i$
$=i+i^2$
$=i-1$
Question:3 If $\frac{(1+i)}{(1-i)}^3-\frac{(1-i)}{(1+i)}^3= x+iy$then find (x, y).
Answer:
Given $\left(\frac{1+i}{1-i}\right)^3-\left(\frac{1-i}{1+i}\right)^3= x+iy$
$=\left(\frac{1+i}{1-i}\right)^3-\left(\frac{1-i}{1+i}\right)^3$
$=\left (\frac{1+2i+i^2}{1-i^2} \right )^{3}-\left (\frac{1-2i+i^2}{1-i^2} \right )^{3}$
$=(\frac{2i}{2})^3-(\frac{-2i}{2})^3$
$=i^3-(-i^3 )$
$=2i^3=0-2i$
$Thus,(x,y)=(0,-2)$
Question:4 If $(1+i)^2/(2-i)= x+iy$ then find the value of x + y.
Answer:
$x+iy=\frac{(1+i)^2}{(2-i)}=\frac{ (1+2i+i^2)}{(2-i)}= \frac{2i}{(2-i)}$
$On\: \: rationalising\: \: 2i(2+i)/(2-i)(2+i)$,
$=\frac{(4i+2i^2)}{(4-i^2 )}=\left (\frac{(4i-2)}{(4+1)} \right )$
$=\frac{-2}{5}+\frac{4i}{5}$
$x=\frac{-2}{5} \: \: and\: \: y=\frac{4}{5}$
$So, x+y= -\frac{2}{5}+\frac{4}{5}=\frac{2}{5}$
Question:5 If $\left(\frac{1-i}{1+i}\right)^{100}=a+ib$ then find (a, b).
Answer:
$a+ib=\left(\frac{1-i}{1+i}\right)^{100}=\left [\frac{1-i}{1+i}*\frac{1-i}{1-i} \right ]^{100}$
$=\left [\frac{(1-i)^2}{1-i^2} \right ]^{100}$
$=\left (\frac{(1-2i+i^2)}{(1+1)} \right )^{100}$
$=\left (\frac {-2i}{2 }\right )^{100}$
$=(i^4 )^{25}=1$
$Hence,(a,b)=(1,0)$
Question:6 If $a=\cos \theta+i \sin\theta$, find the value of$\frac{(1+a)}{(1-a)}$ .
Answer:
$a=\cos \theta+i \sin\theta$
$\frac{(1+a)}{(1-a)}$$=\frac{(1+cos\theta)+i \sin \theta}{(1-cos\theta)-isin\theta}$
$=\frac{\left ( 2\cos ^{2}\frac{\theta }{2} +i2\sin\frac{\theta }{2}\cos\frac{\theta }{2} \right )}{\left ( 2\sin ^{2}\frac{\theta }{2} -i2\sin\frac{\theta }{2}\cos\frac{\theta }{2} \right )}$
$=\frac{2\cos\frac{\theta }{2}\left ( \cos\frac{\theta }{2}+i\sin\frac{\theta }{2} \right )}{2\sin\frac{\theta }{2}\left ( \sin\frac{\theta }{2}-i\cos\frac{\theta }{2} \right )}$
$=\frac{i\cos\frac{\theta }{2}\left ( \cos\frac{\theta }{2}+i\sin\frac{\theta }{2} \right )}{\sin\frac{\theta }{2}\left ( i\sin\frac{\theta }{2}-i^2\cos\frac{\theta }{2} \right )}$
$=\frac{i\cos\frac{\theta }{2}\left ( \cos\frac{\theta }{2}+i\sin\frac{\theta }{2} \right )}{\sin\frac{\theta }{2}\left ( i\sin\frac{\theta }{2}+\cos\frac{\theta }{2} \right )}$
$=i\cot \frac{\theta }{2}$
Question:7 If $(1 + i)z = (1 - i) \bar{z},$ then show that $z= -i\bar{z}$
Answer:
$(1 + i)z = (1 - i) \bar{z},$
$z=\frac{1-i}{1+i}\bar{z}$
Rationalising the denominator,
$\frac{\left (1-i \right )\left (1-i \right )}{\left (1+i \right )\left (1-i \right )}\bar{z}$
$=\frac{\left ( 1-i \right )^{2}}{1-i^2}\bar{z}$
$=\frac{\left ( 1-2i+i^2 \right )}{1+1}\bar{z}$
$=\frac{\left ( 1-2i-1 \right )}{2}\bar{z}$
$=-i\bar{z}$
Hence, Proved.
Answer:
$z=x+iy$
$\bar{z}=x-iy$
$Now,we\: \: also\: \: have\: \: ,z\bar{z }+2(z+\bar{z})+b=0$
$(x+iy)(x-iy)+2(x+iy+x-iy)+b=0$
This equation is an equation of a circle.
Answer:
$Let \: \: z=x+iy \: \:$
$\frac{\bar{z}+2}{\bar{z}-1}= \frac{x-iy+2}{x-iy-1}$
$=\frac{\left [ \left ( x+2 \right )-iy \right ]\left [ \left ( x-1 \right )+iy \right ]}{\left [ \left ( x-1 \right )-iy \right ]\left [ \left ( x-1 \right )+iy \right ]}$
$=\frac{(x-1)(x+2)+y^2+i[(x+2)y-(x-1)y]}{(x-1)^2+y^2}$
$real \: \: part=4 \Rightarrow \frac{(x-1)(x+2)+y^2}{(x-1)^2+y^2}=4$
$x^2+x-2+y^2=4(x^2-2x+1+y^2 )$
$3x^2+3y^2-9x+6=0$
The equation obtained is that of a circle. Hence, the locus of z is a circle.
Answer:
Let z=x+iy
$arg\left ( \frac{z-1}{z+1} \right )=\frac{\pi}{4}$
$arg\left ( z-1 \right )-arg\left ( z+1 \right )=$$\frac{\pi}{4}$
$arg\left ( x+iy-1 \right )-arg\left ( x+iy+1 \right )=$$\frac{\pi}{4}$
$arg\left ( x-1+iy \right )-arg\left ( x+1+iy \right )=$$\frac{\pi}{4}$
$\tan^{-1}\frac{y}{x-1}-\tan^{-1}\frac{y}{x+1}=$$\frac{\pi}{4}$
$\tan^{-1}\frac{\frac{y}{x-1}-\frac{y}{x+1}}{1+\left (\frac{y}{x-1} \right )\left (\frac{y}{x+1} \right )}=$$\frac{\pi}{4}$
$\tan^{-1}\left(\frac{y\left ( x+1+-x+1 \right )}{x^2-1+y^2}\right) = \frac{\pi}{4}$
$\tan\frac{\pi}{4} = \frac{2y}{x^2+y^2-1}$
$x^2+y^2-1=2y$
$x^2+y^2-1-2y=0$
The equation obtained represents the equation of a circle
Question:11 Solve that equation $|z| = z + 1 + 2i$.
Answer:
$|z| = z + 1 + 2i$
$Substituting \: \: z=x+iy \: \: we \: \: get \: \: |x+iy|=x+iy+1+2i$
$|z|= \sqrt{(x^2+y^2 )} =(x+1)+i(y+2)$
Comparing real and imaginary parts $\sqrt{(x^2+y^2 )} =(x+1)$
And$0=y+2 , y=-2$
Substituting the value of y in$\sqrt{(x^2+y^2 )} =(x+1)$
$x^2+(-2)^2=(x+1)^2$
$x^2+4=x^2+2x+1$
Hence, $x=\frac{3}{2}$
Hence,$z=x+iy$
$=\frac{3}{2} -2i$
Question:12 If $|z + 1| = z + 2 (1 + i)$, then find z
Answer:
We have $|z + 1| = z + 2 (1 + i)$
Substituting $z=x+iy$ we get $x+iy+1=x+iy+2i+1$
$|z|= \sqrt{(x^2+y^2 ) }=\sqrt{(x+1)^2+y^2 } = (x+2)+i(y+2)$
Comparing real and imaginary parts, $\sqrt{(x+1)^2+y^2 } = (x+2)$
$and 0=y+2 ;y=-2$
Substituting the value of y in $\sqrt{(x+1)^2+y^2 } = (x+2)$
$(x+1)^2+(-2)^2=(x+2)^2$
$x^2+2x+1+4=x^2+4x+4$
$2x=1$
Hence, $x=\frac{1}{2}$
Thus, $z=x+iy=\frac{1}{2}-2i$
Question:13 If $arg (z - 1) = arg (z + 3i)$, then find x – 1 : y. where $z = x + iy$
Answer:
Given that $arg (z - 1) = arg (z + 3i)$
$arg (x+iy-1)=arg(x+iy+3i)$
$arg(x-1+iy)=arg(x+i(y+3))$
$\tan^{-1}\frac{y}{x-1}=\tan^{-1}\frac{y+3}{x}$
$\frac{y}{x-1}=\frac{y+3}{x}$
$xy=xy-y+3x-3$
$3x-3=y$
$\frac{x-1}{y}=\frac{1}{3}$
Question:14 Show that $\left | \frac{z-2}{z-3} \right |=2$ represents a circle. Find its centre and radius.
Answer:
$\left | \frac{z-2}{z-3} \right |=2$ Substituting $z=x+iy$, we get $\left | \frac{x+iy-2}{x+iy-3} \right |=2$
$|x-2+iy|=2|x-3+iy|$
$\sqrt{(x-2)^2+y^2 }=2\sqrt{((x-3)^2+y^2}$
$x^2-4x+4+y^2=4(x^2-6x+9+y^2 )$
$3x^2+3y^2-20x+32=0$
$x^2+y^2-\frac{20}{3} x+\frac{32}{}3=0$
$(x-\frac{10}{3})^2+y^2+\frac{32}{3}-\frac{100}{9}=0$
Thus, the centre of circle is $(\frac{10}{3},0)$ and radius is $2/3$
Question:15 If $\frac{z-1}{z+1}$ is a purely imaginary number$(z \neq -1)$, then find the value of $|z|$.
Answer:
Let $z=x+iy$
$\frac{z-1}{z+1}=\frac{x+iy-1}{x+iy+1}$
$\frac{\left [ \left ( x-1 \right ) +iy\right ]\left [ \left ( x+1 \right )-iy \right ]}{\left [ \left ( x+1 \right ) +iy\right ]\left [ \left ( x+1 \right )-iy \right ]}$
$=\frac{(x-1)(x+1)+y^2+i[(x+1)y-(x-1)y]}{(x+1)^2+y^2 }$
$\frac{z-1}{z-1}$ is purely imaginary
$\frac{(x-1)(x+1)+y^2}{(x+1)^2+y^2 }=0$
$x^2-1+y^2=0$
$x^2+y^2=1$ Hence, $|z|=1$
Answer:
Let $z_1=\left |z_1 \right |\left (cos\theta _1+i\ sin\theta_1 \right )$ and $z_2=\left |z_2 \right |\left (cos\theta _2+i\ sin\theta_2 \right )$
Given that $|z_1| = |z_2|$
And $arg\left (z_1 \right )+arg\left (z_2 \right )= \pi$
$\theta_1+\theta_2=\pi$
$\theta_1=\pi-\theta_2$
$z_2=\left |z_2 \right |\left (cos\theta _2+i\ sin\theta_2 \right )$
$z_1=|z_2 |\left ( -\cos \theta_2+isin\theta_2 \right )$
$z_1=-|z_2 |\left ( \cos \theta_2-isin\theta_2 \right )$
$z_1=-|z_2 |\left ( \cos \theta_2-isin\theta_2 \right )$, $|z_1| = -|z_2|$
Question:17 If |z1| = 1 (z1 ≠ –1) and $z_2= \frac{z_1-1}{z_1+1}$ then show that the real part of z2 is zero
Answer:
Let $z_1=x+iy$ $|z_1 |= \sqrt{x^2+y^2} =1$
$z_2= \frac{z_1-1}{z_1+1}=\frac{x+iy-1}{x+iy+1}$
$=\frac{[(x-1)+iy][(x+1)-iy]}{[(x+1)+iy][(x+1)-iy] }$
$= \frac{(x-1)(x+1)+y^2+i[(x+1)y-(x-1)y]}{(x+1)^2+y^2}$
$\frac{x^2+y^2-1+2iy}{(x+1)^2+y^2 }=0$
Since, $x^2+y^2=1$
$\frac{1-1+2iy}{(x+1)^2+y^2}=\frac{0+2iy}{(x+1)^2+y^2}$
Therefore, the real part of $z_2$ is zero
Answer:
$z_1$ and $z_2$ are conjugate complex numbers.
The negative side of the real axis $= r_1 (cos\theta-isin \theta)$
$=r_1\left ( \cos\left ( -\theta_1 \right )+i\sin\left (- \theta_1 \right ) \right )$
Similarly, $z_3=r_2\left ( \cos\left ( \theta_2 \right )-i\sin\left ( \theta_2 \right ) \right )$
$z_4=r_2\left ( \cos\left ( -\theta_2 \right )+i\sin\left (- \theta_2 \right ) \right )$
$arg \left ( \frac{z_1}{z_4} \right )+arg \left ( \frac{z_2}{z_3} \right )$$=arg(z_1)-arg(z_4)+ arg(z_2)- arg(z_3)??$
$=\theta_1-(-\theta_2)+(-\theta_1)-\theta_2=0$
Answer:
$|z_1| = |z_2| =... = |z_n| = 1$
$|z_1|^2 = |z_2|^2 =... = |z_n|^2 = 1$
$z_1 \bar{z_1 } =z_2 \bar{z_2 }=...=z_n \bar{z_n }$
$z_1=\frac{1}{\bar{z_1}},z_2=\frac{1}{\bar{z_2}},...z_n=\frac{1}{\bar{z_n}}$
Now, $|z_1+z_2+z_3+...z_n |$ $=\left | \frac{z_1\bar{z_1}}{\bar{z_1}}+\frac{z_2\bar{z_2}}{\bar{z_2}}+...+\frac{z_n\bar{z_n}}{\bar{z_n}} \right |$
=$\left | \frac{1}{\bar{z_1}}+\frac{1}{\bar{z_2}}+...+\frac{1}{\bar{z_n}} \right |$
$=\left | {\frac{1}{z_1}+\frac{1}{z_2}+...+\frac{1}{z_n}} \right |$
Answer:
Let $z_1=\left | z_1 \right |\left ( \cos\theta_1 +i\sin \theta_1 \right )$ and $z_2=\left | z_2 \right |\left ( \cos\theta_2 +i\sin \theta_2 \right )$
$arg(z_1 )- arg(z_1 )=0$
$\theta _1-\theta _2=0$
$\theta _1=\theta _2$
$z_1-z_2= (z_1 |cos\theta_1-| z_2|cos\theta_1 )+i(z_1 |sin\theta_1-| z_2|sin\theta_1 )$
$z_1-z_2= \sqrt{(z_1 |cos\theta_1-| z_2|cos\theta_1 )^{2}+(z_1 |sin\theta_1-| z_2|sin\theta_1 )^{2}}$
$=\sqrt{|z_1|^2+|z_2|^2-2|z_1||z_2|\cos^2\theta_1-2|z_1||z_2|\sin^2\theta_1}$
$=\sqrt{|z_1|^2+|z_2|^2-2|z_1||z_2|}$
$=\sqrt{\left (|z_1|-|z_2| \right )^2}= |z_1|-|z_2|$
Question:21 Solve the system of equations $Re (z^2) = 0, |z| = 2.$
Answer:
$Re (z^2) = 0, |z| = 2.$
$Let\: \: z=x+iy$
$|z|= \sqrt{x^2+y^2 } =2$
${x^2+y^2 } =4$
$z^2=x^2+2ixy-y^2$
$=(x^2-y^2 )+2ixy$
$Now,Re (z^2 )=0$
$x^2-y^2=0$
$x^2=y^2=2$
$x=y=\pm \sqrt{2}$
Hence, $z=x+iy=\pm\sqrt{2}\pm i\sqrt{2}$
$=\sqrt{2}+i\sqrt{2},\sqrt{2}-i\sqrt{2},-\sqrt{2}+i\sqrt{2} \: \: and\: \: -\sqrt{2}-i\sqrt{2}$
Question:22 Find the complex number satisfying the equation $z+\sqrt{2} |(z+1)|+i=0$.
Answer:
$z+\sqrt{2} |(z+1)|+i=0\: \: \: \: ...(i)$
Substituting $z=x+iy$ we get $x+iy+\sqrt2 |x+iy+1|+i=0$
$x+i(1+y)+\sqrt2 \left [ \sqrt{(x+1)^2+y^2 } \right ]=0$
$x+i(1+y)+\sqrt2 \left [ \sqrt{(x^2+2x+1+y^2 ) } \right ]=0$
1+y = 0
y = -1
$x+\sqrt2 \sqrt{x^2+2x+2}=0$
$\sqrt2 \sqrt{x^2+2x+2}=-x$
$2x^2+4x+4=x^2$
$x^2+4x+4=0$
$(x+2)^2=0$
$x=-2$
Hence,$z=x+iy=-2-i$
Question:23 Write the complex number $z=\frac{1-i}{\cos \frac{\pi}{3}+i\sin \frac{\pi}{3}}$ in polar form
Answer:
$z=\frac{1-i}{\cos \frac{\pi}{3}+i\sin \frac{\pi}{3}}$
$=\frac{\sqrt{2}\left [ \frac{1}{\sqrt{2}}-i1/\sqrt{2} \right ]}{\cos \frac{\pi}{3}+i\sin \frac{\pi}{3}}$
$=\frac{\sqrt{2}\left [\cos\frac{\pi}{4}-isin\frac{\pi}{4} \right ]}{\cos \frac{\pi}{3}+i\sin \frac{\pi}{3}}$
$=\sqrt2\left [ \cos \left ( -\frac{\pi}{4}-\frac{\pi}{3}\right )+i\sin\left (-\frac{\pi}{4}-\frac{\pi}{3} \right ) \right ]$
$=\sqrt2\left [ \cos \left ( -\frac{7\pi}{12}\right )+i\sin\left (-\frac{7\pi}{12} \right ) \right ]$
$=-\sqrt2\left [ \cos \left ( \frac{5\pi}{12}\right )+i\sin\left (\frac{5\pi}{12} \right ) \right ]$
Answer:
Let $z=\left | z \right |(\cos\theta_1 + i sin\theta_1 )$ and $w=\left | w \right |(\cos\theta_2 + i sin\theta_2 )$
$|zw|=|z||w|=1$
Also $arg(z)-arg(w)= \frac{\pi}{2}$
$\theta_1-\theta_2=\frac{\pi}{2}$
$\bar{z} w=|z|\left (\cos \theta _{1}-i\sin\theta _{1} \right )w$
$w=|w|\left (\cos \theta _{2}+i\sin\theta _{2} \right )=1$
$\bar{z}w$$=|z||w|(\cos (-\theta_1)+i sin(-\theta_1 ))(cos\theta_2+i \sin\theta_2 )$
$\bar{z}w$$=\cos [(\theta _2 - \theta _1 )+isin( \theta _2- \theta _1 )]$
$=\left [ \cos \left ( -\frac{\pi}{2} \right ) +i\sin\left ( -\frac{\pi}{2} \right )\right ]$
$=1[0-i]=-i$
Answer:
i) $|az_1 - bz_2|^2 + |bz_1 + az_2|^2$
$=|az_1 |^2+|bz_2 |^2-2Re(az_1.b(\bar{z_2} ) +|bz_1 |^2+|az_2 |^2+2Re(az_1.b\bar{z_2} ) ?)$
$=|az_1 |^2+|bz_2 |^2+|bz_1 |^2+|az_2 |^2=(a^2+b^2 )(|z_1 |^2+|z_2 |^2 )$
ii)$\sqrt{-25}*\sqrt{-9}=5i*3i=15i^2=-15$
iii) $\frac{(1-i)^3}{1-i^3}=\frac{(1-i)^3}{(1-i)(1+i+i^2 )}$
$=\frac{(1-i)^3}{(1+i-1) }= \frac{1+i^2-2i}{}i$
$=\frac{1-1-2i}{i}=-\frac{2i}{i}=-2$
iv) $i+i^2+i^3+...+i^{1000}=0 \left [ \sum_{n=1}^{1000}i^n=0 \right ]$
v) $\frac{1}{1+i}=1*\frac{1-i}{(1+i)(1-i)} =\frac{1}{2} (1-i)$
vi) Let $z_1=x_1+iy_1 \: \: and \: \: z_2=x_2+iy_2$
$z_1+z_2=(x_1+x_2 )+i(y_1+y_2 )$
If $z_1+z_2$ is real then $y_1+y_2=0$
$y_1=-y_2$
$z_2=x_2-iy_1$
$z_2=x_1-iy_1 (\: \: when\: \: x_1=x_2 )$
So, $z_2=\bar{z_1}$
vii) $arg (z)+arg (\bar{z} )$
$If\: \: arg (z)= \theta , then \arg (\bar{z})=-\theta$
$\theta+(-\theta)=0$
viii) $|z+4|\leq 3$
$=|z+4-3|\leq|z+4|+|-3|$
$=|z+4-3|\leq3+3$
$=|z+4-3|\leq6$
The greatest value is 6 and the least value of $|z+1|$ is 0
ix)$\frac{z-2}{z+2}=\frac{\pi}{6}$
Let $z=x+iy$
$\left |\frac{x+iy-2}{x+iy+2} \right |=\left |\frac{(x-2)+iy}{(x+2)+iy} \right |= \frac{\pi}{6}$
$6|(x-2)+iy|=\pi|(x+2)+iy|$
$6\sqrt{(x-2)^2+y^2 }=\pi\sqrt{(x+2)^2+y^2 }$
$36[x^2+4-4x+y^2 ]=\pi ^2[x^2+4+4x+y^2 ]$
$36x^2+144-144x+36y^2=\pi^2 x^2+4\pi^2+4\pi^2 x+\pi^2 y^2$
$(36-\pi^2 ) x^2+(36-\pi^2 )-(144+4\pi^2 )x+144-4\pi^2=0$
Which represents an equation of a circle
x) $|z|=4$ and $arg (z) = \frac{5\pi}{6}$
Let $z=x+iy$
$|z|= \sqrt{x^2+y^2} =4$
$x^2+y^2=16$
$arg(z)=\tan^{-1}\frac{y}{x}=\frac{5\pi}{6}$
$\frac{y}{x}=\tan\left ( \pi-\frac{\pi}{6} \right )$
$=-\tan \frac{\pi}{6}=-\frac{1}{\sqrt{3}}$
$x=-\sqrt3 y$
$(-\sqrt3 y)^2+y^2=16$
$3y^2+y^2=16$
$4y^2=16$
$y^2=4$
$y=\pm2$
$x=-2\sqrt3$
$z=-2\sqrt3+2i$
Answer:
(i) Comparison of two purely imaginary complex numbers is not possible. However, the two purely real complex numbers can be compared. So, it is false.
(ii) Let $z= x+iy$
$z.i = (x+iy)i =xi= -y$ which rotates at angle of 180. So, it is ‘false’.
(iii) Let $z= x+iy$
$|z|+|z-1|= \sqrt{x^2+y^2} +\sqrt{(x-1)^2+y^2}$
The value of $|z|+|z-1|$ is minimum when $x=0,y=0 \, \: \: i.e.,1$
Hence, it is true.
iv) Let $z= x+iy$
Given that $\left | z-1 \right |=\left |z-i \right |$
$|x+iy-1|=|x+iy-i|$
$|(x-1)+iy|=|x-(1-y)i|$
$\sqrt{(x-1)^2+y^2}=\sqrt{x^2+(1-y)}^2$
$(x-1)^2+y^2=x^2+(1-y)^2$
$x^2-2x+1+y^2=x^2+1+y^2-2y$
$-2x+2y=0 x-y=0$ which is a straight line slope=1
Now, the equation of the line through the point 1,0and 0,1
$y-0=\frac{1-0}{0-1} (x-1)$
$y=-x+1$ whose slope=-1
Multiplication of the slopes of two lines =-1*1=-1
So, they are perpendicular. Hence, it is true.
v)Let $z= x+iy$ $z\neq0 \: \: and \: \: Re(z)=0$
Since, the real part is 0 $x=0, z=0+iy=iy$
$1m(z^2 )=y^2 i^2=-y^2$which is real Hence, it is False.
vi) $(z-4)<|z-2|$
Let $z= x+iy$
$|x+iy-4|<|x+iy-2|$
$|(x-4)+iy|<|(x-2)+iy|$
$\sqrt{(x-4)^2+y^2 }<\sqrt{(x-2)^2+y^2}$
$(x-4)^2+y^2<(x-2)^2+y^2$
$(x-4)^2<(x-2)$
$x^2+16-8x<x^2+4-4x$
$8x+4x<-16+4$
$-4x<-12$
$x>3$ Hence, it is true.
vii) Let $z_1=x_1+iy_1$ and $z_2=x_2+iy_2$
$|z_1+z_2 |=|z_1 |+|z_2 |$
$|x_1+iy_1+x_2+iy_2 |=|(x_1+iy_1 )+(x_2+y_2 i)|$
$= \sqrt{(x_1+x_2 )^2+(y_1+y_2 )^2 }$
$=\sqrt{x_1^2+y_1^2} +\sqrt{x_2^2+y_2^2 }$
Squaring both sides, we get $(x_1^2+x_2^2+2x_1 x_2+y_1^2+y_2^2+2y_1 y_2$)
$=x_1^2+y_1^2+x_2^2+y_2^2+2\sqrt{x_1^2 x_2^2+x_1^2 y_2^2+x_2^2 y_1^2+y_1^2 y_2^2 }$
$=2x_1 x_2+2y_1 y_2=\sqrt{x_1^2 x_2^2+x_1^2 y_2^2+x_2^2 y_1^2+y_1^2 y_2^2 }$
$=x_1 x_2+y_1 y_2=\sqrt{x_1^2 x_2^2+x_1^2 y_2^2+x_2^2 y_1^2+y_1^2 y_2^2 }$
Again squaring on both sides we get $x_1^2 x_2^2+x_1^2 y_2^2+2x_1 y_1 x_2 y_2=x_1^2 x_2^2+x_1^2 y_2^2+x_2^2 y_1^2+y_1^2 y_2^2 2x_1 y_1 x_2 y_2=x_1^2 y_2^2+x_2^2 y_1^2$
$x_1^2 y_1^2+x_2^2 y_1^2-2x_1 y_1 x_2 y_2=0$
$(x_1 y_2-x_2 y_1 )^2=0$
$x_1 y_2-x_2 y_1=0$
$x_1 y_2=x_2 y_1$
$\frac{x_1}{y_1}=\frac{x_2}{y_2}$
$\frac{y_1}{x_1}=\frac{y_2}{x_2}$
$arg(z_1)=arg(z_2)=0$ Hence, it is false.
(viii) Since every real number is a complex number. So, 2 is a complex number. Hence, it is false.
Question:27
a) The polar form of $i+\sqrt3$ is | i) Perpendicular bisector of the segment joining (-2,0) and (2,0) |
b)The amplitude of $-1+\sqrt{-3}$ is | ii) On or outside the circle having centre at (0,-4) and radius 3. |
c) If $\left | z+2 \right |=\left | z-2 \right |$ then locus of z is | iii) $2\pi/3$ |
d)If $\left | z+2i \right |=\left | z-2i \right |$ then locus of z is | iv)Perpendicular bisector of the segment joining (0,-2) and (0,2) |
e) Region represented by $\left | z+4i \right |\geq 3$ is | v)$2\left ( \cos\frac{\pi}{6}+i\sin\frac{\pi}{6} \right )$ |
f) Region represented by $\left | z+4 \right |\geq 3$ is | vi) On or outside the circle having a centre at (-4,0) and a radius of 3 units. |
g)Conjugate of $\frac{1+2i}{1-i}$ lies in | vii) First Quadrant |
h) Reciprocal of 1-i lies in | viii) Third Quadrant |
Answer:
a)Given $z=i+\sqrt3$
Polar form of z $=r[cos\theta+isin\theta]=i+\sqrt3$.
$r=\sqrt{3+1} =2$
And $\tan \alpha = \frac{1}{\sqrt3}$
$\alpha = \frac{\pi}{6}$
Since $x>0,y>0$
Polar form of z $=2[cos\frac{\pi}{6}+isin\frac{\pi}{6}]$
b) Given that $z=-1+\sqrt3=-1+\sqrt3 i$
Here argument z $=\tan^{-1}\left | \frac{\sqrt3}{-1} \right |= \tan^{-1}\sqrt{3}= \frac{\pi}{3}$
So, $\alpha = \frac{\pi}{3}$
Since, x<0 and y>0 $\theta=\pi-\alpha=\pi-\frac{\pi}{3}=\frac{2\pi}{3}$
c) $|z+2|=|z-2|$
$|x+iy+2|=|x+iy-2|$
$|(x-2)+iy|=|(x-2)+iy|$
$\sqrt{(x+2)^2+y^2} =\sqrt{(x-2)^2+y^2}$
$(x+2)^2+y^2=(x-2)^2+y^2$
$(x+2)^2=(x-2)^2$
$x^2+4+4x=x^2+4-4x$
$8x=0 , x=0$
which represents the equation of the y-axis and is perpendicular to the line joining the points (-2,0) and (2,0)
d) $|z+2i|=|z-2i|$
$(x+iy+2i)=|x+iy-2i|$
$|x+(y+2)i|=|x+(y-2)i|$
$\sqrt{x^2+(y+2)^2} =\sqrt{x^2+(y-2)^2 }$
$x^2+(y+2)^2=x^2+(y-2)^2$
$(y+2)^2=(y-2)^2$
$y^2+4+4y=y^2+4-4y$
$8y=0 ,y=0$
which is the equation of the x-axis and is perpendicular to the line segment joining (0,-2)and (0,2)
e) $|z+4i|\geq3$
$|x+iy+4|\geq3$
$|x+(y+4)i|\geq3$
$\sqrt{x^2+(y+4)^2} \geq3$
$x^2+(y+4)^2\geq9$
$x^2+y^2+8y+16 \geq9$
$x^2+y^2+8y+7 \geq0$ $r=\sqrt{(4)^2-7}=3$
which represents a circle on or outside having a centre (0,-4)
f) $|z+4|\leq3$
Let $z=x+iy$ Then, $|x+iy+4|\leq3$
$|(x+4)+iy|\leq3$
$\sqrt{(x+4)^2+y^2 } \leq3$
$x^2+8x+16+y^2\leq9$
$x^2+y^2+8x+7 \leq0$ which is a circle having centre (-4,0) and $r=\sqrt{(4)^2-7}=\sqrt9=3$ and is on the circle.
g) Let $z=\frac{1+2i}{1-i}$
$\frac{1+2i}{1-i}* \frac{1+i}{1+i}=\frac{1+i+2i+2i^2}{1-i^2}$
$=\frac{1+i+2i-2}{1+1}= \frac{-1+3i}{2}$
$=\frac{1+i+2i-2}{1+1}= \frac{-1+3i}{2}$
$=-\frac{1}{2}+\frac{3}{2}i$ which lies in third quadrant..
h) Given that z=1-i
Reciprocal of $z=\frac{1}{z}=\frac{1}{1-i}*\frac{1+i}{1+i}$
$=\frac{1+i}{1+i^2}$ which lies in first quadrant.
$So,(a)-(v),(b)-(iii),(c)-(i),(d)-(iv),(e)-(ii),(f)-(vi),(g)-(viii),(h)-(vii)$
Question:28 What is the conjugate of $\frac{2-i}{\left (1-2i \right )^2}$?
Answer:
$\frac{2-i}{\left (1-2i \right )^2}$$= \frac{2-i}{1+4i^2-4i}$
$= \frac{2-i}{-3-4i}= \frac{2-i}{-3-4i}*\frac{-3+4i}{-3+4i}$
$=\frac{-6+8i+3i-4i^2}{9-16i^2}$
$\frac{-6+11i+4}{9-16i^2}=\frac{-2+11i}{9+16}$
$\frac{-2+11i}{25}=\frac{-2}{25}+\frac{11}{25}i$
Hence, $\bar{z}=-\frac{2}{25}-\frac{11}{25}i$
Question:29 If $|z_1| = |z_2|$, is it necessary that $z_1 = z_2$?
Answer:
Let $z_1=x_1+iy_1 \: \: and\: \: z_2=x_2+iy_2$
$|x_1+iy_1 |=|x_2+iy_2 |$
$= \sqrt{x_1^2+y_1^2} = \sqrt{x_2^2+y_2^2}$
$x_1^2+y_1^2=x_2^2+y_2^2$
$\\$Consider $ z_1 = 3 + 4i \\ z_2 = 4 + 3i$
$z_1\neq z_2$ Hence, it is not necessary that $z_1= z_2$
Question:31 Find z if $|z| = 4$ and $arg (z) = 5\pi/6$..
Answer:
Polar form of $z=r[cos \theta+i\sin \theta]$
$=4\left [ \cos \frac{5\pi}{6}+i\sin \frac{5\pi}{6} \right ]$
$=4\left [ \cos\left ( \pi-\frac{\pi}{6} \right )+i\sin \left ( \pi-\frac{\pi}{6} \right ) \right ]$
$=4\left [ -\cos\left (\frac{\pi}{6} \right )+i\sin \left (\frac{\pi}{6} \right ) \right ]$
$=4\left [-\frac{\sqrt3}{2}+i\frac{1}{2} \right ]$
$=-2\sqrt3+2i$
Question:32 Find $(1+i)\frac{(2+i)}{(3+i)}$.
Answer:
$\left | (1+i)\frac{(2+i)}{(3+i)}*\frac{3-i}{3-i} \right |=\left | (1+i).\frac{6-2i+3i-i^2}{9-i^2} \right |$
$=\left | \frac{(1+i)(7+i)}{9+1} \right |$
$=\left | \frac{7+i+7i+i^2}{9+1} \right |$
$=\left | \frac{7+i+7i+i^2}{10} \right |$
$=\left | \frac{7+8i-1}{10} \right |$
$=\left | \frac{6+8i}{10} \right |$
$=\left |\frac{3}{5}+\frac{4}{5}i \right |$
$=\sqrt{\left (\frac{3}{5} \right )^2+\left (\frac{4}{5} \right )^2}=1$
Question:33 Find the principal argument of $(1+i\sqrt3)^2$.
Answer:
$\begin{array}{c} (1+i \sqrt{3})^{2}=1+i^{2} \cdot 3+2 \sqrt{3} i \\ =1-3+2 \sqrt{3} i=-2+2 \sqrt{3} i \\ \tan \alpha=\left|\frac{2 \sqrt{3}}{-2}\right| \\ \tan \alpha=|-\sqrt{3}|=\sqrt{3} \\ \tan \alpha=\tan \frac{\pi}{3} \\ \alpha=\frac{\pi}{3} \end{array}$
Now Re(z)<0 and image(z)>0 $arg(z)=\pi-\alpha=\pi-\frac{\pi}{3}=\frac{2\pi}{3}$
$\text {Hence, the principal arg}=\frac{2 \pi}{3}$
Question:34 Where does z lie, if $|\frac{z-5i}{z+5i}|=1$.
Answer:
$\begin{array}{c} \text { Let } z=x+i y \quad\left|\frac{x+y i-5 i}{x+i y+5 i}\right|=1 \\ |x+(y-5) i|=|x+(y+5) i| \\ x^{2}+(y-5)^{2}=x^{2}+(y+5)^{2} \\ (y-5)^{2}=(y+5)^{2} \\ \qquad \begin{array}{r} y^{2}+25-10 y=y^{2}+25+10 y \\ 20 y=0 \quad y=0 \end{array} \end{array}$
Hence, z lies on the x-axis i.e., the real axis.
Answer:
The answer is option (c).
let $z=\sin x+i \cos 2 x \quad \bar{z}=\sin x-i \cos 2 x$
But we are given that $\bar{z}=\cos x-i \sin 2 x$
$\sin x-i \cos 2 x=\cos x-i \sin 2 x$
Comparing the real and imaginary parts, we get
$\sin x=\cos x \quad$ and $\quad \cos 2 x=\sin 2 x$
$\tan \mathrm{x}=1 \text { and } \tan 2 \mathrm{x}=1$
$\tan 2 \mathrm{x}= \frac{2\tan x }{1-\tan^2 x}=1$
The above value is not satisfied by tan x = 1.
Hence, no value of x is possible.
Answer:
The answer is the option (c).
Let $z=\frac{\left ( 1-i\sin\alpha \right )}{1+2i\sin\alpha}$
$\begin{aligned} &\begin{array}{c} =\frac{(1-i\sin \alpha)(1-2 i \sin \alpha)}{(1+2 i \sin \alpha)(1-2 i \sin \alpha)} \\\\ =\frac{1-2 i \sin \alpha-i \sin \alpha+2 i^{2} \sin ^{2} \alpha}{(1)^{2}-(2 i \sin \alpha)^{2}} \\\\ =\frac{1-3 i \sin \alpha-2 \sin ^{2} \alpha}{1-4 i^{2} \sin ^{2} \alpha}=\frac{\left(1-2 \sin ^{2} \alpha\right)-3 i \sin \alpha}{1+4 \sin ^{2} \alpha} \\\\ \frac{-3 i \sin \alpha}{1+4 \sin ^{2} \alpha}=0 \\\\ \sin \alpha=0 \quad \operatorname{} \alpha=n \pi \end{array}\\ &\text { Hence, c is correct. } \end{aligned}$
Answer:
The answer is the option (b).
If z lies in the third quadrant,So,$x<0 \: \: and\: \: y<0\: \: , z=x+iy$
$\begin{aligned} \frac{\bar{z}}{z} &=\frac{x-i y}{x+i y}=\frac{x-i y}{x+i y} * \frac{x-i y}{x-i y} \\ &=\frac{x^{2}+i^{2} y^{2}-2 x y i}{x^{2}-i^{2} y^{2}} \\ &=\frac{x^{2}-y^{2}}{x^{2}+y^{2}}-\frac{2 x y}{x^{2}+y^{2}} i \end{aligned}$
When z lies in the third quadrant then $\frac{\bar{z}}{z}$ will also be in the third quadrant.
$\begin{array}{c} \frac{x^{2}-y^{2}}{x^{2}+y^{2}}<0 \text { and } \frac{2 x y}{x^{2}+y^{2}}>0 \\\\ x^{2}-y^{2}<0 \text { and } 2 x y>0 \\\\ x^{2}<y^{2} \text { and } x y>0 \\\\ \text { So, } x<y<0 \end{array}$
Hence, b is correct.
Question:38 The value of $(z + 3) (\bar{z}+3)$ is equivalent to A.$|z + 3|^2$ B. |z – 3| C.$z^2 + 3$ D. None of these
Answer:
The answer is the option (a).
$Let\: \: z=x+iy \: \: So,\: \: (z+3)(\bar{z} +3)=(x+iy+3)(x-iy+3)$
$=\left [ \left ( x+3 \right ) +iy\right ]\left [ \left ( x+3 \right ) -iy\right ]$
$=(x+3)^2-y^2 i^2=(x+3)^2+y^2$
$=|x+3+iy|^2=|z+3|^2$
Question:39 If $\left ( \frac{1+i}{1-i} \right )^{x}=1$ , then A. $x = 2n + 1$ B. $x = 4n$ C. $x = 2n$ D.$x = 4n + 1$
Answer:
The answer is the option(b).
$\begin{array}{c} \left(\frac{1+i}{1-i}\right)^{x}=1 \\\\ \left(\frac{(1+i)(1+i)}{(1-i)(1+i)}\right)^{x}=1 \\\\ \left(\frac{1-1+2 i}{1+1}\right)^{x}=1 \\\\ \left(\frac{2 i}{2}\right)^{x}=1(i)^{x}=1 \end{array}$
$x = 4 n, n \epsilon N$
Hence, the correct option is
Answer:
The answer is the option (a).
$\begin{aligned} &\text {Given that} \frac{3-4 i x}{3+4 i x}=\alpha-i \beta\\ &=\frac{3-4 i x}{3+4 i x} * \frac{3-4 i x}{3-4 i x}=\alpha-i \beta\\ &\frac{9-12 i x-12 i x+16 i^{2} x^{2}}{9-16 i^{2} x^{2}}=\alpha-i \beta\\ &=\frac{9-16 x^{2}}{9+16 x^{2}}-\frac{24 x}{9+16 x^{2}}=\alpha-i \beta \ldots \ldots .(i)\\ &\frac{9-16 x^{2}}{9+16 x^{2}}+\frac{24 x}{9+16 x^{2}}=\alpha+i \beta \ldots \ldots \ldots(ii)\\\end{aligned}$
Multiplying eqn (i )and (ii )we get
$\left (\frac{9-16 x^{2}}{9+16 x^{2}} \right )^{2}+\left (\frac{24 x}{9+16 x^{2}} \right )^{2}=\alpha^{2}+ \beta^{2}$
$\frac{81+256 x^{4}-288 x^{2}+576x^2}{\left(9+16 x^{2}\right)^{2}}=\alpha^{2}+\beta^{2}$
$\begin{aligned}\frac{81+256 x^{4}+288 x^{2}}{\left(9+16 x^{2}\right)^{2}}=\alpha^{2}+\beta^{2}\\ \frac{\left(9+16 x^{2}\right)^{2}}{\left(9+16 x^{2}\right)^{2}}=\alpha^{2}+\beta^{2}=1 \end{aligned}$
Answer:
The answer is the option (a).
$\begin{array}{c} z_{1}=r_{1}\left(\cos \theta_{1}+i \sin \theta_{2}\right) \\ \left|z_{1}\right|=r_{1} \\ z_{2}=r_{2}\left(\cos \theta_{2}+i \sin \theta_{2}\right) \\ \left|z_{2}\right|=r_{2} \\\\ z_1z_2=r_{1} r_{2}\left[\cos \left(\theta_{1}+\theta_{2}\right)+i \sin \left(\theta_{1}+\theta_{2}\right)\right] \\ =r_{1}\left(\cos \theta_{1}+i \sin \theta_{1}\right) \cdot r_{2}\left(\cos \theta_{2}+i \sin \theta_{2}\right) \\ \end{array}$
$\begin{array}{c} =r_{1} r_{2}\left[\left(\cos \theta_{1} \cos \theta_{2}-\sin \theta_{1} \sin \theta_{2}\right)+i\left(\sin \theta_{1} \cos \theta_{2}+\cos \theta_{1} \sin \theta_{2}\right)\right] \\ =r_{1} r_{2}\left[\cos \left(\theta_{1}+\theta_{2}\right)+i \sin \left(\theta_{1}+\theta_{2}\right)\right]\left|z_{1} z_{2}\right|=\left|z_{1} \| z_{2}\right| \end{array}$
Hence, option A is correct.
Answer:
The answer is the option (b).
If z rotated through an angle of $\pi/2$ about the origin in a clockwise direction.
Then the new position $=z.e^{-\left ( \frac{\pi}{2} \right )}$
$=\left (2-i \right ).e^{-\left ( \frac{\pi}{2} \right )}$
$=\left (2-i \right ).\left [ \cos\left (- \frac{\pi}{2} \right )+isin\left (- \frac{\pi}{2} \right ) \right ]$
$=\left (2-i \right ).\left (0-i \right )$
$=-1-2i$
Hence, the correct option is (b)
Answer:
The answer is the option (d).
$x+iy$ is a non-real complex number if $y\neq0$,
Hence, d is correct
Answer:
The answer is the option (d).
$a+ib=c+id$
$|a+ib|=|c+id|$
$\sqrt{a^2+b^2} =\sqrt{c^2+d^2}$
Squaring both sides, $a^2+b^2=c^2+d^2$
Hence, d is correct
Answer:
The answer is the option (b).
$\begin{array}{c} \text { Let } z=x+i y \quad\left|\frac{i+x+i y}{i-x-y i}\right|=1 \\\\ \left|\frac{x-(y+1) i}{-x-(y-1) i}\right|=1 \\\\ |x+(y+1) i|=|-x-(y-1) i| \\\\ \sqrt{x^{2}+(y+1)^{2}}=\sqrt{x^{2}+(y-1)^{2}} \\\\ x^{2}+(y+1)^{2}=x^{2}+(y-1)^{2} \\\\ y^{2}+2 y+1=y^{2}-2 y+1 \\ 2 y=-2 y \\\\ y=0 ; x-a x i s \end{array}$
Hence, b is correct
Answer:
. The answer is the option (b).
$Let \: \: z=x+iy \: \: \: \: ,|z|^2=x^2+y^2$
$Now,z^2=x^2+y^2 i^2+2xyi$
$z^2=x^2-y^2+2xyi$
$|z|^2=\sqrt{(x^2-y^2 )^2+(2xy)^2 }$
$=\sqrt{x^4+y^4+2x^2 y^2} =\sqrt{(x^2+y^2 )^2}$
$|z|^2=x^2+y^2=|z|^2$
Hence, b is correct
Answer:
The answer is the option (c).
Since, $|z_1+z_2 |=|z_1 |+|z_2 |$
$|z_1+z_2 |=|r_1 \left ( \cos \theta _1+i\sin\theta _1 \right )+r_2 \left ( \cos \theta _2+i\sin\theta _2 \right )|$
$=\sqrt{r_{1}^{2} \cos ^{2} \theta_{1}+r_{2}^{2} \cos ^{2} \theta_{2}+2 r_{1} r_{2} \cos \theta_{1} \cos \theta_{2}+r_{1}^{2} \sin ^{2} \theta_{1}+r_{2}^{2} \sin ^{2} \theta_{2}+2 r_{1} r_{2} \sin \theta_{1} \sin \theta_{2}}$
So, $\begin{array}{c} =\sqrt{r_{1}^{2}+r_{2}^{2}+2 r_{1} r_{2} \cos \left(\theta_{1}-\theta_{2}\right)} \\ \left|z_{1}+z_{2}\right|=\left|z_{1}\right|+\left|z_{2}\right| \\\\ \sqrt{r_{1}^{2}+r_{2}^{2}+2 r_{1} r_{2} \cos \left(\theta_{1}-\theta_{2}\right)}=r_{1}+r_{2} \end{array}$
Squaring both sides,
$\begin{array}{c} r_{1}^{2}+r_{2}^{2}+2 r_{1} r_{2} \cos \left(\theta_{1}-\theta_{2}\right)=r_{1}^{2}+r_{2}^{2}+2 r_{1} r_{2} \\ 2 r_{1} r_{2}-2 r_{1} r_{2} \cos \left(\theta_{1}-\theta_{2}\right)=0 \\ 1-\cos \left(\theta_{1}-\theta_{2}\right)=0 \\ \cos \left(\theta_{1}-\theta_{2}\right)=1 \\ \theta_{1}-\theta_{2}=0 \\ \theta_{1}=\theta_{2} \\ \operatorname{so} \arg \left(z_{1}\right)=\arg \left(z_{2}\right) \end{array}$
Answer:
Let $z=\frac{1+i \cos \theta}{1-2 i \cos \theta}=\frac{1+i \cos \theta}{1-2 i \cos \theta} * \frac{1+2 i \cos \theta}{1+2 i \cos \theta}$
On solving, we get
$\frac{1+3 i \cos \theta-2 \cos ^{2} \theta}{1-4 i^{2} \cos ^{2} \theta} \\=\frac{1+3 i \cos \theta-2 \cos ^{2} \theta}{1+4 \cos ^{2} \theta}$
$=\frac{\left(1-2 \cos ^{2} \theta\right)+3 i \cos \theta}{1+4 \cos ^{2} \theta}$
If z is a real number, then
$\frac{3 \cos \theta}{1+4 \cos ^{2} \theta}=0$
$\\3 \cos \theta=0 \\\cos \theta=0$
$\theta=\frac{(2 n+1) \pi}{2}, n \in N$
Question:49 The value of $arg (x)$ when $x < 0$is: A. 0 B. $\frac{\pi}{2}$ C. π D. none of these
Answer:
The answer is option (c).
$\\ Let\: \: z= -x+0i \: \: and \: \: x<0 \\ |z|=\sqrt{(-1)^2+(0)^2 }=1,x<0$
Since, the point (-x,0)lies on the negative side of the real axis, principal argument (z)=π
Hence, option c is correct.
Answer:
The answer is option (a).
Given that $\\ z=1+2i \\ |z|=\sqrt{(1)^2+(2)^2 }=\sqrt5$
Now $f(z)=\frac{7-z}{1-z^2}$
$\begin{aligned} =\frac{7-(1+2 i)}{1-(1+2 i)^{2}} &=\frac{7-1-2 i}{1-1-4 i^{2}-4 i} \\ =\frac{6-2 i}{4-4 i} &=\frac{3-i}{2-2 i} \end{aligned}$
$\begin{aligned} =\frac{3-i}{2-2 i} * \frac{2+2 i}{2+2 i}=\frac{6+6 i-2 i-2 i^{2}}{4-4 i} \\ =\frac{6+4 i+2}{4+4} &=\frac{8+4 i}{8} \\ =1+\frac{1}{2} i & \end{aligned}$
$f(z)=\sqrt{(1)^{2}+\left(\frac{1}{2}\right)^{2}}=\sqrt{1+\frac{1}{4}}=\frac{\sqrt{5}}{2}=\frac{|z|}{2}$
Hence, option A is correct.
Various laws need to be followed and derived by a mathematician when it comes to multiplying two complex numbers. All of them are named after the features of the specific formula derivation. Ahead, you will also learn about the Modulus and Conjugate of complex numbers and how to solve sums with their help. There is an Argand and Polar representation in the chapter, relating to graphical presentations. A point needs to be derived at the ‘x’ and ‘y’ axis to form an Argand or complex plane.
Students, with the help of NCERT Exemplar Class 11 Maths Solutions Chapter 5, will not face any issues when trying to solve such problems. It will also help them score better in exams.
All the concepts have been covered in the NCERT Exemplar Solutions for Class 11 Maths chapter 5. By using the NCERT Exemplar Class 11 Maths chapter 5 solutions PDF Download function, students can access quality study material that is effectively constructed by experts for the best learning experience.
Complex numbers and quadratic equations are not only useful chapters in class 11 but also for higher studies and competitive exams. Strengthening basic concepts is a necessity for students so that later they do not face any difficulties solving Complex numbers and quadratic equations questions in higher studies or competitive examinations.
Some important facts about solving Complex numbers and quadratic equations in class 11 are listed below.
Students can study strategically at their own pace after accessing Class 11 Maths NCERT Solutions chapter 5. This will boost their confidence to attempt other questions from this chapter.
Class 11 Maths Chapter 5 NCERT solutions are solved by subject-matter experts and are very reliable at the same time. The solutions provide shortcuts as well as detailed explanations with necessary formulae that will help students to understand the answers better.
These solutions will help students manage their time efficiently in this chapter and understand which questions are easier to approach and which are time-consuming. This will be helpful during the exam.
NCERT solutions for class 11 Maths chapter 5 Complex numbers and quadratic equations is designed to give the students step-by-step solutions for a particular question.
Read more NCERT Solution subject-wise
Also, read NCERT Notes subject-wise -
Also, check NCERT Books and NCERT Syllabus here:
Here are some useful links for NCERT books and the NCERT syllabus for class 11:
The Argand Plane is significant in Class 11 Maths Chapter 5 (Complex Numbers) as it provides a geometric representation of complex numbers. Each complex number z=x+iy is plotted as a point (x,y) in a 2D plane, where the horizontal axis represents the real part and the vertical axis represents the imaginary part. This visual approach helps understand complex number operations like addition, subtraction, modulus, and conjugation more intuitively.
The conjugate of a complex number z=x+i y is . It reflects the number across the real axis on the Argand Plane and is useful in division and simplification.
The reciprocal of $z$ is given by:
This expression helps divide complex numbers by rationalizing the denominator using the conjugate.
Important topics covered in NCERT Exemplar Class 11 Maths Chapter 5 (Complex Numbers and Quadratic Equations) include:
Algebra of complex numbers (addition, subtraction, multiplication, division)
Modulus and argument of complex numbers
Polar and exponential form
Geometrical representation on Argand Plane
Solving quadratic equations with complex roots
Use of identities and inequalities involving complex numbers
To find the modulus and argument of a complex number z=x+i y :
- Modulus is the distance from the origin:
- Argument is the angle $\theta$ made with the positive real axis:
The modulus shows the length, and the argument shows the direction of the complex number on the Argand Plane.
In NCERT Exemplar Class 11 Maths, the polar form of a complex number expresses z=x+i y in terms of its modulus and argument
. It is written as:
This form is useful for multiplication, division, and finding powers/roots of complex numbers. It provides a clear geometric interpretation using the Argand Plane.
Take Aakash iACST and get instant scholarship on coaching programs.
This ebook serves as a valuable study guide for NEET 2025 exam.
This e-book offers NEET PYQ and serves as an indispensable NEET study material.
As per latest syllabus. Physics formulas, equations, & laws of class 11 & 12th chapters
As per latest syllabus. Chemistry formulas, equations, & laws of class 11 & 12th chapters
As per latest 2024 syllabus. Study 40% syllabus and score upto 100% marks in JEE